Skip to main content

Extra Credit Blog!!!!

WHAT WOULD HAPPEN IF EVERYONE JUMPED??

If everyone in the world (7.6 billion people) gathered in one place and all jumped at the same time, what would happen to the Earth? Would it be effected? 


This is a question that scientists have been asking for years. The answer simply put is that even if everyone in the world jumped at once, the Earth would virtually be unaffected. This is because the mass of the earth is approximately 5.972 x 10^24 kg, and all human mass put together would not even come close to that number. To see the exact number of how the Earth would be effected, I used the conservation of momentum equation. 

p1i + p2i = p1f + p2f
m1v1+m2v2=m1v1f+m2v2f

I know that in this situation the equation is elastic because the Earth is in the system and momentum is always conserved. In all my calculations, 1 will represent the Earth and 2 will represent the people. 

First here are the numbers I used for the data:
  • Average human mass = 136 lbs / 62 kg (<https://www.livescience.com/36470-human-population-weight.html>)
  • Human population as of December 2017 = 7.6 billion
  • Mass of the Earth = 5.972 x 10^24 kg 
  • Average vertical jump = 0.5m
Because this is an elastic collision, this means that energy is also conserved. I started this problem using a conservation of energy equation to solve for the velocity of the Earth. This I showed by just using the initial point before the jump when delta KE = delta PE. The Earth did not move up or down, so h = 0.

1/2m1v1^2 + 1/2m2v2^2 = mgh
h initial = 0 
1/2m1v1^2 + 1/2m2v2^2 = 0
1/2m1v1^2 = -1/2m2v2^2
m1v1= - m2v2
v1 = -m2v2/m1

I then used the same energy equation to calculate the velocity of the people, when h did not = 0. The mass in the potential energy is the m2: the mass of the people. 


1/2m1v1^2 + 1/2m2v2^2 = m2gh
m1v1^2 + m2v2^2 = 2m2gh
m2v2^2 = 2m2gh-m1v1^2
v2^2 = (2m2gh-m1v1^2)/m2
v2^2 = 2gh - m1v1^2/m2
v2 = sq. rt. 2gh - m1v1^2/m2

I then put the two equations together to solve for the velocity of the Earth. Here was the result:

v1^2 = (2ghm2^2/m1^2) - (m2v1^2/m1)

From here I had to solve for v1 to get the velocity of the Earth.

v1^2 + (m2v1^2/m1) = (2ghm2^2/m1^2)
v1^2 (1 +m2/m1) = (2ghm2^2/m1^2)
v1^2 = 2ghm2^2 / m1^2 (1 +m2/m1)
v1^2 = 2ghm2^2/ (m1^2+m2m1)
v1 = sq. rt. 2ghm2^2/ (m1^2+m2m1)

Now that I found this equation, all I had to do was plug in the values for each thing that I mentioned above. 
First I found the mass of the total population of the people which was (7.6 x 10^9)*62 = 4.712 x 10^11 kg. 

v2 = sq. rt. 2*9.8* (4.712 x 10^11)^2 / (5.972 x 10^24)^2 + (5.972 x 10^24)*(4.712 x 10^11)
v2 = (2.17589 x 10^24)/ (3.56648 x 10^49)
v2 = 6.10095 x 10^-26 m/s

This means that the velocity of the Earth after the jump would only be effected by 6.10095 x 10^-26 m/s. Virtually, there would be no effect on the Earth. This is probably because the total mass of the human population is so small compared to the Earth it would be hard to effect the Earth's momentum. The larger the mass, the harder it is to move. This means in order for the human population to change the Earth's orbit they would need a giant number for their velocity in order to make a dent. 

So at the end of the day, if everyone decided to gather in the same place and jump it would probably just be a waste of their time. 

Comments

Post a Comment

Popular posts from this blog

The Physics of Spiderman

Over this past weekend after I finished working on my homework, I decided to relax and watch a few movies before going asleep. Among the movies I watched was Spider-Man 3 from 2007 and despite the movie flaws I was interested by the scenes that showed Spider Man shooting through the sky with the use of his webs that come out of his wrists. Due to this, I decided to make my blog post about the physics of Spider-Man's slingshot. After doing some research, I discovered just how much information there is on the physics of Spider-Man and how elements of Spider-Man can be used as examples for most topics learned in mechanics. For this investigation, I will not be using the horrible cliche and terrible CGI infested mess that Spider-Man 3 is but instead the all around superior Spider-Man movie of Spider-Man 2 to investigate the physics of Spider-Man's web propelled slingshot.  I want to talk about what happens in terms of physics when Spider-Man launches himself across a dista...
Post a Comment
Read more

Physics of Black Holes...Or Lack Thereof

Isabella Jacavone To comprehend how the universe works, we must dwell into the most basic building blocks of existence; matter, energy, space, and time. NASA's  Physics of the Cosmos program involves cosmology, astrophysics, and fundamental physics intended to answer questions about the elusiveness of complex concepts such as black holes, neutron stars, dark energy, and gravitational waves. In this blog post, I'd like to elaborate on a subject that is very intriguing  to me; Black holes. And more specifically, what would happen if we got near one. A black hole is anything but a hole, but rather an immense amount of matter compacted into an extremely small area. A black hole is caused when, hypothetically, a star four times more massive than our sun collapses into a sphere no bigger than 600 square km. To put that in perspective, that's about the size of New York City. B lack holes were predicted by Einstein's theory of general relativity, which showed that when a...
Post a Comment
Read more

Circular and Projectile Motion in The Real World: Hammer Throw

Hammer Throw, along with other track and field events, were one of the first sports to partake in the Olympics. The idea of hammer throw and other field events is rather simple; for hammer, a 7.26 kg metal ball is attached to a 3 foot long steel wire and the thrower has to make the metal ball travel as far as possible. Olympic and professional hammer throwers have perfected the sport down to a science - and science it is. To make the projectile travel as far as possible, the thrower must spin. Throwing shoes are worn to decrease traction and the hammer is outstretched while travelling in an orbital pattern as the thrower spins.  According to   Brian LeRoy , University of Arizona associate professor of physics, taking home the gold in the hammer throw requires three things from athletes: throw as close to 45 degrees as possible, spin faster, and keep arms fully extended. LeRoy says, theoretically, throwing the hammer at a 45-degree angle is ideal. Gra...
Post a Comment
Read more